Question

The radius of a sphere is increasing at the rate of 2 m/sec. find the rate of change of the volume with respect to time when radius is 4m?

Answer 1

Sphere Volume is given by `V=4/3pir^3` where r is given in meters for example (whichever linear unit can be used)

Radius varies in function of time `r=2t`

If radius increase at rate of `2m/sec` in t seconds it will be `2m/(sec) t sec=2t` meters in t seconds

In our volume equation `V=4/3pi(2t)^3=32/3pit^3`

The volume varies each t seconds acording to this formula

Answer 2

`sf(128picolor(white)(x)m^3"/s")`

Explanation:

The volume of a sphere is given by:

`sf(V=4/(3)pir^3)`

We are told that:

`sf((dr)/(dt)=2color(white)(x)"m/s")`

Differentiating implicitely with respect to t::

`sf((dV)/(dt)=4/(3)pi.3r^2.(dr)/dt)`

When `sf(r=4color(white)(x)m)` this becomes:

`sf((dV)/(dt)=4/(cancel(3))picancel(3)xx16xx2=128picolor(white)(x)m^3"/s")`